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Find the number of trailing zeros in the product of

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by kullayappayenugula » Tue Nov 13, 2012 10:50 pm
(1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50).

if you observe the number are multiples of 5 and are power of the same number.

therefore the expression is as

(1^1)*(5^5)*(10^10)*(15^15)*(20^20)*(25^25)*(35^35)*(40^40)*(45^45)*(50^50).
=> (1^1)*(5^5)*(10^10)*(3^15*5^15)*(2^20*10^20)............(5^50*10^50)

now all the ten terms that can be formed are
10^10*10^20*10^30*10^40*10^50*(5^100*2^100)

=>10^250
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by FLUID » Tue Nov 13, 2012 11:21 pm
nasheen wrote:Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50).

A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

C

We have a trailing zero when we 2 is multiplied by 5. In the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there are more 5-s than 2-s so the number of 2's will be deciding factor for the number of trailing zeros.

If we factor 2's then it is 10 ^ 10, 20 ^ 20, 30 ^ 30, 40 ^ 40, 50 ^ 50
= 2 ^ (10+40+30+120+50) * (Remaining) = 2 ^ (250) * (Remaining)

So there will be 250 trailing zero
Thanks,

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by nasheen » Wed Nov 14, 2012 6:28 am
FLUID wrote:
nasheen wrote:Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50).

A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

C



We have a trailing zero when we 2 is multiplied by 5. In the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there are more 5-s than 2-s so the number of 2's will be deciding factor for the number of trailing zeros.

If we factor 2's then it is 10 ^ 10, 20 ^ 20, 30 ^ 30, 40 ^ 40, 50 ^ 50
= 2 ^ (10+40+30+120+50) * (Remaining) = 2 ^ (250) * (Remaining)

So there will be 250 trailing zero
I dint got these concept of trailing zero
Why we need to multiply 2 by 5?
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